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The minimum-degree condition $\delta(G)\ge n/2$ implies Ore's degree-sum condition

Statement

Let $G$ be a finite simple graph on $n\ge1$ vertices. If $\delta(G)\ge n/2$ , then every nonadjacent pair $u,v$ satisfies $\deg(u)+\deg(v)\ge n$ .

Facts & Assumptions

Given: A nonempty finite simple graph $G$ on $n$ vertices with $\delta(G)\ge n/2$ .

[F1]

The minimum degree satisfies $\delta(G)\le\deg(w)$ for every vertex $w$ (Adjacency, incidence, open and closed neighbourhoods, vertex degree, minimum degree and maximum degree).

[F2]

The order $n$ is the finite cardinality of the vertex set (The cardinality $\lvert A\rvert$ of a finite set).

Proof

technique · direct

1.1

For any nonadjacent vertices $u,v$ , [F1] and the hypothesis give $\deg(u)+\deg(v)\ge2\delta(G)\ge n$ .

givenF1F2algebra

2.1

Since the pair was arbitrary, Ore's degree-sum condition holds for every nonadjacent pair.

step 1.1∎

Depends on

Used by

Dirac's theorem: every n-vertex graph with n≥3 and δ(G)≥ n/2 is Hamiltonian Corollary

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 26 results over 13 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

Applied Combinatorics, Eulerian and Hamiltonian Graphs (standard reference, not scraped)

The minimum-degree condition δ(G)≥n/2\delta(G)\ge n/2δ(G)≥n/2 implies Ore's degree-sum condition